1. ## Normal Distribution

Supose we have a random sample of 2n from a population denoted by X and E(X) = μ and V(X) = σ2. Let X1 bar = (1/2n) ∑i=12n Xi and X2 bar = (1/n) ∑i=1n Xi be two estimators of μ. Which is a better estimator of μ? explain why

i believe its the latter because This is the better estimator because it is an unbiased estimator of the mean but any additional suggestions and input would be useful..

2. They're both unbiased. They are both sample means. But the first one has more observations hence it has a smaller variance. Thus it is preferable.
Let m=n and also 2n.
$E(\bar X)=E\biggl({X_1+\cdots +X_m\over m}\biggr)={1\over m}\biggl(E(X_1)+\cdots +E(X_m)\biggr)={m\mu\over m}=\mu$
IN BOTH cases. So they are both unbiased.
NOW using independence we can likewise obtain
$V(\bar X)={V(X_1)\over m}={\sigma^2\over m}$.
Hence the winner is the one with a larger sample size, which is the first one.
AND, there is no need for normality here whatsoever.

3. Originally Posted by HoJo1001

Supose we have a random sample of 2n from a population denoted by X and E(X) = μ and V(X) = σ2. Let X1 bar = (1/2n) ∑i=12n Xi and X2 bar = (1/n) ∑i=1n Xi be two estimators of μ. Which is a better estimator of μ? explain why

i believe its the latter because This is the better estimator because it is an unbiased estimator of the mean but any additional suggestions and input would be useful..
Your choice of estimators for the mean of a data set of size $2n$ are $\overline{X}_1 = \frac{1}{2n} \sum_{i=1}^{2n} x_i$ and $\overline{X}_2 = \frac{1}{n} \sum_{i=1}^{n} x_i$.

Surely $\overline{X}_1$ is better since it uses more data!

Unless there is some special way you haven't mentioned for selecting the n data used to calculate $\overline{X}_2$.